##n^2## is divided by 3, then n is also divided by 3.

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Homework Statement


Prove that if ##n^2## is divided by 3, then also n can also be divided by 3.

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The Attempt at a Solution



Is there anything wrong with this argument?

By contradiction, suppose ##3|n## and 3 doesn't divide ##n^2##.
Then ##3m = n##. Multiple both sides by ##n##
$$
3(mn) = n^2
$$
Thus, ##3|n^2## and we have reached a contradiction. Therefore, If ##3|n^2##, then ##3|n##.
 
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Right. If n=3m then n2=9m2 which is always divisible by 3.
 
Dustinsfl said:
By contradiction, suppose 3|n3|n and 3 doesn't divide n2n^2.
This is not right. Your prove that the conjecture 3|n Λ not 3|n^2 produces a contradiction, but that proves that 3|n ⇒ 3|n^2.
You need to prove that 3|n^2 ⇒ 3|n
 
What you have proved is that "if n is divisible by 3 then n^2 is divisible by 3. That's the "wrong way around"!
To give a proof by contradiction, you need to start "suppose n is NOT divisible by 3", not that is divisible by 3.

Do you see that if n is not divisible by 3, it must be of the form "3k+ 1" or "3k+ 2" for some integer k?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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